If a⃗ = α î + β ĵ + 3k̂,
b⃗ = - β î - α ĵ - k̂ and
c⃗ = î - 2ĵ - k̂
such that a⃗ . b⃗ = 1 and b⃗ . c⃗ = - 3, then 1/3( ( a⃗ × b⃗ ) . c⃗ ) is equal to _____________.

Solution

Maths sir · 4 steps

Use the two dot products to pin down α and β first, then the triple product is just a 3 by 3 determinant.

1. Use a·b = 1. a⃗ · b⃗ = -αβ - αβ - 3 = -2αβ - 3. Setting this equal to 1 gives αβ = -2.
2. Use b·c = −3. b⃗ · c⃗ = -β + 2α + 1 = -3, so β = 2α + 4. Substituting into αβ = -2 gives 2α² + 4α + 2 = 0, that is (α + 1)² = 0. So α = -1 and β = 2.
3. Write the determinant. Now a⃗ = (-1, 2, 3), b⃗ = (-2, 1, -1) and c⃗ = (1, -2, -1). The triple product (a⃗ × b⃗) · c⃗ is the determinant with these as rows.

   |-1, 2, 3; -2, 1, -1; 1, -2, -1| = -1(-1-2) - 2(2+1) + 3(4-1) = 3 - 6 + 9 = 6

4. So. 1/3((a⃗ × b⃗) · c⃗) = 6/3 = 2.

The common slip: 6. Stopping at the determinant and forgetting the factor 1/3 gives 6, not 2.

The idea to keep. Unknown components plus given dot products means solve for the unknowns first. Then (a⃗ × b⃗) · c⃗ is the determinant of the three vectors written as rows.

Answer from the NTA final answer key. Solution by PrepWiser.